Show that if $lim_{xto c}(f(x))^2=0$ then $lim_{xto c}f(x)=0$












1












$begingroup$



Let $cinmathbb{R}$ and let $f:mathbb{R}tomathbb{R}$ be such that
$$lim_{xto c}(f(x))^2=L$$
Show that if $L=0$, then
$$lim_{xto c}f(x)=0$$




Using the $delta$ - $varepsilon$ definition, it is given that:

$forall varepsilon>0$ $existsdelta_1>0$ such that, if $|x-c|<delta_1$ then, $|f^2(x)|<varepsilon$


Using this, what we need to show is that:

$forall varepsilon>0$ $existsdelta_2>0$ such that, if $|x-c|<delta_2$ then, $|f(x)|<varepsilon$




My approach:


Since, $varepsilon$ is taken to be small, we can say that $varepsilon<1$. I take $delta_2=delta_1$ $forallvarepsilon>0$.
$Rightarrow|f^2(x)|<|f(x)|<varepsilon<1$, which doesn't get me anywhere.


(In fact, this makes it look like that the converse is true!)




My approach is obviously very wrong. How should I actually proceed?






real-analysis limits epsilon-delta







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Let $cinmathbb{R}$ and let $f:mathbb{R}tomathbb{R}$ be such that
    $$lim_{xto c}(f(x))^2=L$$
    Show that if $L=0$, then
    $$lim_{xto c}f(x)=0$$




    Using the $delta$ - $varepsilon$ definition, it is given that:

    $forall varepsilon>0$ $existsdelta_1>0$ such that, if $|x-c|<delta_1$ then, $|f^2(x)|<varepsilon$


    Using this, what we need to show is that:

    $forall varepsilon>0$ $existsdelta_2>0$ such that, if $|x-c|<delta_2$ then, $|f(x)|<varepsilon$




    My approach:


    Since, $varepsilon$ is taken to be small, we can say that $varepsilon<1$. I take $delta_2=delta_1$ $forallvarepsilon>0$.
    $Rightarrow|f^2(x)|<|f(x)|<varepsilon<1$, which doesn't get me anywhere.


    (In fact, this makes it look like that the converse is true!)




    My approach is obviously very wrong. How should I actually proceed?






    real-analysis limits epsilon-delta







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Let $cinmathbb{R}$ and let $f:mathbb{R}tomathbb{R}$ be such that
      $$lim_{xto c}(f(x))^2=L$$
      Show that if $L=0$, then
      $$lim_{xto c}f(x)=0$$




      Using the $delta$ - $varepsilon$ definition, it is given that:

      $forall varepsilon>0$ $existsdelta_1>0$ such that, if $|x-c|<delta_1$ then, $|f^2(x)|<varepsilon$


      Using this, what we need to show is that:

      $forall varepsilon>0$ $existsdelta_2>0$ such that, if $|x-c|<delta_2$ then, $|f(x)|<varepsilon$




      My approach:


      Since, $varepsilon$ is taken to be small, we can say that $varepsilon<1$. I take $delta_2=delta_1$ $forallvarepsilon>0$.
      $Rightarrow|f^2(x)|<|f(x)|<varepsilon<1$, which doesn't get me anywhere.


      (In fact, this makes it look like that the converse is true!)




      My approach is obviously very wrong. How should I actually proceed?






      real-analysis limits epsilon-delta







      share|cite|improve this question











      $endgroup$





      Let $cinmathbb{R}$ and let $f:mathbb{R}tomathbb{R}$ be such that
      $$lim_{xto c}(f(x))^2=L$$
      Show that if $L=0$, then
      $$lim_{xto c}f(x)=0$$




      Using the $delta$ - $varepsilon$ definition, it is given that:

      $forall varepsilon>0$ $existsdelta_1>0$ such that, if $|x-c|<delta_1$ then, $|f^2(x)|<varepsilon$


      Using this, what we need to show is that:

      $forall varepsilon>0$ $existsdelta_2>0$ such that, if $|x-c|<delta_2$ then, $|f(x)|<varepsilon$




      My approach:


      Since, $varepsilon$ is taken to be small, we can say that $varepsilon<1$. I take $delta_2=delta_1$ $forallvarepsilon>0$.
      $Rightarrow|f^2(x)|<|f(x)|<varepsilon<1$, which doesn't get me anywhere.


      (In fact, this makes it look like that the converse is true!)




      My approach is obviously very wrong. How should I actually proceed?






      real-analysis limits epsilon-delta




      real-analysis limits epsilon-delta






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 11:00









      José Carlos Santos

      173k23133241




      173k23133241










      asked Dec 20 '18 at 10:44









      Za IraZa Ira

      161115




      161115






















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          $begingroup$

          Let $varepsilon>0$. You know that there is a $delta>0$ such that $$lvert xrvert<deltaimpliesbigllvert f^2(x)bigrrvert<varepsilon^2.$$But$$bigllvert f^2(x)bigrrvert<varepsilon^2iffbigllvert f(x)bigrrvert<varepsilon.$$






          share|cite|improve this answer









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            1 Answer
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            active

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            $begingroup$

            Let $varepsilon>0$. You know that there is a $delta>0$ such that $$lvert xrvert<deltaimpliesbigllvert f^2(x)bigrrvert<varepsilon^2.$$But$$bigllvert f^2(x)bigrrvert<varepsilon^2iffbigllvert f(x)bigrrvert<varepsilon.$$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Let $varepsilon>0$. You know that there is a $delta>0$ such that $$lvert xrvert<deltaimpliesbigllvert f^2(x)bigrrvert<varepsilon^2.$$But$$bigllvert f^2(x)bigrrvert<varepsilon^2iffbigllvert f(x)bigrrvert<varepsilon.$$






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Let $varepsilon>0$. You know that there is a $delta>0$ such that $$lvert xrvert<deltaimpliesbigllvert f^2(x)bigrrvert<varepsilon^2.$$But$$bigllvert f^2(x)bigrrvert<varepsilon^2iffbigllvert f(x)bigrrvert<varepsilon.$$






                share|cite|improve this answer









                $endgroup$



                Let $varepsilon>0$. You know that there is a $delta>0$ such that $$lvert xrvert<deltaimpliesbigllvert f^2(x)bigrrvert<varepsilon^2.$$But$$bigllvert f^2(x)bigrrvert<varepsilon^2iffbigllvert f(x)bigrrvert<varepsilon.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 20 '18 at 10:51









                José Carlos SantosJosé Carlos Santos

                173k23133241




                173k23133241






























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